Problem: Evaluate the following expression. Your answer must be exact. $\left(2-2\sqrt{3}i\right)^3=$
The Strategy The easiest way to find $z^{n}$ for a complex number $z=({a}+{b}i)$ is using its modulus and argument. Therefore, our solution will consist of the following steps: Find the modulus and argument of $z$. [How is this done, in general?] Find the modulus and argument of $z^{n}$. [How is this done, in general?] Find the rectangular form $z^{n}$. Find the modulus and argument of $\left(2-2\sqrt{3}i\right)$ $({2}{-2\sqrt{3}}i)$ is of the form $({a}+{b}i)$, where ${a=2}$ and ${b=-2\sqrt{3}}$. Therefore: $\begin{aligned}r&=\sqrt{{a}^2 + {b}^2} \\\\&=\sqrt{({2})^2 + ({-2\sqrt{3}})^2} \\\\&=\sqrt{{4}+{12}} \\\\&=4\end{aligned}$ Using the arctangent formula, we have: $\begin{aligned}\theta&=\arctan\left(\dfrac{{b}}{{a}}\right) \\\\&=\arctan\left(\dfrac{{-2\sqrt{3}}}{{2}}\right) \\\\&=-60^\circ\end{aligned}$ Since ${a=2}$ is positive and ${b=-2\sqrt{3}}$ is negative, $(2-2\sqrt{3}i)$ lies in Quadrant $4$. Therefore, $\theta$ must be between $270^\circ$ and $360^\circ$. Using the identity $\tan(360+\theta)=\tan(\theta)$, we know that the following is also a solution of the equation. $360^\circ+(-60^\circ)=300^\circ$ So $\theta = 300^{\circ}$. Find the modulus and argument of $\left(2-2\sqrt{3}i\right)^3$ We found that the modulus and argument of $({2}{-2\sqrt{3}}i)$ are $4$ and $300^\circ$. Therefore, the modulus and argument of $({2}{-2\sqrt{3}}i)^3$ are $4^3=64$ and $(300^\circ)\cdot3=900^\circ=180^\circ$. Find the rectangular form of $\left(2-2\sqrt{3}i\right)^3$ Since the argument is $180°$, we know the number lies on the negative side of the real number axis and is therefore a negative real number. Since the modulus is $64$, our solution is $-64$. [What does this look like graphically?] [How do we find this algebraically?] Summary $\left(2-2\sqrt{3}i\right)^3=-64$